regex match for startswith

Question

Given a random length of filepath, how to regex for something in the middle:

name = 'path/to/../../foo/file.py'

something like this:

In [2]: name.startswith('.*/foo/')
Out[2]: False

Have you looked into using [`re`](https://docs.python.org/2/library/re.html) ? — Nir Alfasi, Nov 17 '15 at 02:21
Check this: http://stackoverflow.com/questions/33747968/getting-file-list-using-glob-in-python/33748006 — Remi Guan, Nov 17 '15 at 02:38

Ravichandra · Answer 1 · 2015-11-17T03:46:49.660

0

Instead of using re Use glob as the way to do it.

import glob
for filename in glob.iglob('C:\Users\Desktop\**\Start\c*'):
    print(filename)

output

C:\Users\Desktop\screenshoots\Start\cmr(1).PNG
C:\Users\Desktop\screenshoots\Start\cmr(2).PNG
C:\Users\Desktop\screenshoots\Start\cmr(3).PNG

edited Nov 17 '15 at 03:46

answered Nov 17 '15 at 02:26

Ravichandra

1 Answers1